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I know that increasing subshell decreases ionization energy; for example, going from Beryllium to Boron. I do not understand why this is.

The answer my textbook (Chemistry 2e on OpenStax) gives is that the higher subshells (eg $\ce{p}$) has more energy than the lower subshells (eg $\ce{s}$). I do not understand why a higher energy subshell require more energy to remove (ie more ionization energy)? Is it as if the higher energy subshell is less stable, and thus has lower ionization energy?

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    $\begingroup$ Yeah, that doesn't make sense. Ionization energy does increase in going from Be to B, but it's not because you're going to a higher-energy subshell (which you are), it's because you're staying in the same shell but increasing the effective nuclear charge. Could you provide the exact quote from your textbook? $\endgroup$
    – theorist
    Aug 4, 2021 at 2:48
  • $\begingroup$ 2p electrons have higher energy ( and lower ionization energy ) than 2s electrons within the same atom context, but these energies shift in the different element atom context. So 2s/2p( if excited ) electrons in Be have higher energies ( lower ionization energies ) than respective 2s/2p electrons in B. $\endgroup$
    – Poutnik
    Aug 4, 2021 at 8:04

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I think in order to understand this phenomenon you have to consider what is happening during an ionization. During this process we are removing a single electron from our molecule or atom. The electron we are going to remove will be the one which is easiest to remove (the one we require the least amount of energy to remove).

Having this in mind, we can take the knowledge that the electron shells keep increasing in energy so e.g. 2s < 2p. "Higher in energy" means that they are less bound to the nucleus (that's just the way we usually define our energy scale).

Thus when combining these two, we can understand why the ionization energy decreases when ionizing atoms that have electrons in higher shells.

The probably more interesting question here is probably why the energy of the shells keeps increasing. To break it down the reason is that electrons in higher shells have (on average) a larger distance to the nucleus. As the electrostatic potential (which is is the main interaction between electron and nucleus) is proportional to the inverse distance (search for "Coulomb potential" if you want more details), this means that these electrons just feel less and less of the attraction from the nucleus which causes them to be more easily removed.

Another effect that plays into this is called "shielding" and it essentially means that the electrons from the lower shells will shield the nucleus' positive charge to some degree and thus the outer electrons the ones in the higher shells) will feel less attractive force to begin with.

Note that when comparing different atoms you also have to take into account that these have different nuclei with different charges so the comparison between atoms (in contrast to different shell within a single atom) can be a bit more complicated than this view alone and might require to crunch the actual numbers but I think conceptually this can help explaining the trends we are seeing throughout the PSE.

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